Mathematical Induction
Serial order wise

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Prove 1 + 2 + 3 + ……. + n = (𝐧(𝐧+𝟏))/𝟐 for n, n is a natural number Step 1: Let P(n) : (the given statement) Let P(n): 1 + 2 + 3 + ……. + n = (n(n + 1))/2 Step 2: Prove for n = 1 For n = 1, L.H.S = 1 R.H.S = (𝑛(𝑛 + 1))/2 = (1(1 + 1))/2 = (1 × 2)/2 = 1 Since, L.H.S. = R.H.S ∴ P(n) is true for n = 1 Step 3: Assume P(k) to be true and then prove P(k + 1) is true Assume that P(k) is true, P(k): 1 + 2 + 3 + ……. + k = (𝑘(𝑘 + 1))/2 We will prove that P(k + 1) is true. P(k + 1): 1 + 2 + 3 +……. + (k + 1) = ((k + 1)( (k + 1) + 1))/2 P(k + 1): 1 + 2 + 3 +…….+ k + (k + 1) = ((𝐤 + 𝟏)(𝐤 + 𝟐))/𝟐 We have to prove P(k + 1) is true Solving LHS 1 + 2 + 3 +…….+ k + (k + 1) From (1): 1 + 2 + 3 + ……. + k = (𝑘(𝑘 + 1))/2 = (𝒌(𝒌 + 𝟏))/𝟐 + (k + 1) = (𝑘(𝑘 + 1) + 2(𝑘 + 1))/2 = ((𝒌 + 𝟏)(𝒌 + 𝟐))/𝟐 = RHS ∴ P(k + 1) is true when P(k) is true Step 4: Write the following line Thus, By the principle of mathematical induction, P(n) is true for n, where n is a natural number

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo